ISS 2016 Statistics Paper-2 Solution: Question 41 (Unbiased Estimator of Uniform Distribution)

Continuing our question-by-question walk through the ISS Statistics Paper-2 previous year papers, we now arrive at Question 41 of the 2016 paper. This one sits squarely in the estimation theory portion of the syllabus, and it is a question that rewards candidates who are comfortable going back to the basic definition of expectation rather than memorising a formula.

Quick Summary

• Topic: Unbiased estimation for the Uniform distribution U(0, θ)

• Question reference: ISS 2016, Statistics Paper-2, Question 41

• Correct answer: Option (c) — 2X̄

The Question as Asked

If X₁, X₂, X₃, ..., Xₙ follows U(0, θ) be a random sample for i = 1, 2, 3, ..., n, then what is an unbiased estimator of θ?

• (a) X̄

• (b) X̄/2

• (c) 2X̄

• (d) √X̄

Step 1: Write Down What We Are Given

We are told that each Xᵢ is drawn independently from a Uniform distribution on the interval (0, θ). Its probability density function is

f(x; θ) = 1/θ, for 0 < x < θ, and 0 otherwise.

An estimator T is called unbiased for a parameter θ if, on average over repeated sampling, it equals θ exactly — that is, E(T) = θ. So the entire problem reduces to one task: compute the expectation of each candidate statistic and see which one equals θ.

Step 2: Find E(Xᵢ) From First Principles

Since every Xᵢ has the same distribution, we only need the expectation of a single Uniform(0, θ) random variable. By definition,

E(X) = ∫₀^θ x · f(x) dx = ∫₀^θ x · (1/θ) dx

Pull the constant 1/θ outside the integral:

E(X) = (1/θ) ∫₀^θ x dx

Now integrate x with respect to x. Recall ∫x dx = x²/2, so:

∫₀^θ x dx = [x²/2]₀^θ = θ²/2 − 0 = θ²/2

Substitute this back:

E(X) = (1/θ) · (θ²/2) = θ/2

So every single observation Xᵢ has expectation θ/2, not θ. This is the fact the entire question hinges on.

Step 3: Find E(X̄)

The sample mean is X̄ = (X₁ + X₂ + ... + Xₙ)/n. Using linearity of expectation (the expectation of a sum is the sum of expectations, and constants pull straight through):

E(X̄) = E[(X₁ + X₂ + ... + Xₙ)/n] = (1/n)[E(X₁) + E(X₂) + ... + E(Xₙ)]

Since each E(Xᵢ) = θ/2, and there are n of them:

E(X̄) = (1/n) · n · (θ/2) = θ/2

So the sample mean X̄, by itself, has expectation θ/2 — it systematically undershoots θ by a factor of 2, on average. That immediately rules out option (a).

Step 4: Check Each Option Against E(X̄) = θ/2

Now we just scale correctly. If E(X̄) = θ/2, we want a multiple of X̄ whose expectation is exactly θ.

Option (a): X̄.E(X̄) = θ/2 ≠ θ. Biased (too small by half).

Option (b): X̄/2.Using the property E(cT) = c·E(T) for a constant c, E(X̄/2) = (1/2)·E(X̄) = (1/2)·(θ/2) = θ/4 ≠ θ. Biased, and even further off than option (a).

Option (c): 2X̄.E(2X̄) = 2·E(X̄) = 2·(θ/2) = θ. This matches exactly. Unbiased.

Option (d): √X̄.This is a nonlinear function of X̄, and expectation does not pass through square roots the way it does through multiplication by a constant — in general E(√X̄) ≠ √E(X̄), and by Jensen's inequality for a concave function like the square root, E(√X̄) is actually strictly less than √(θ/2) itself, let alone θ. So this option is also biased, and there isn't even a clean closed form for it.

Pro Tip: Whenever a problem gives you a distribution and asks you to find an unbiased estimator, your very first move should always be to compute E(single observation), not to stare at the answer options. Once you have E(Xᵢ) in terms of θ, every option becomes a one-line check using E(cT) = c·E(T). Students who drill this compute-E(Xᵢ)-first-then-scale habit until it is automatic rarely lose marks on estimation questions, because it turns a four-option guessing game into pure arithmetic.

Step 5: Final Answer and Confirmation

Putting it together: E(2X̄) = θ, so 2X̄ is the unbiased estimator of θ among the given options. The correct choice is option (c), and this agrees exactly with the official ISS 2016 Statistics Paper-2 answer key, which also lists (c) as the answer for Question 41.

Why This Question Matters

This question is a clean test of whether a candidate actually understands the definition of unbiasedness rather than having memorised a table of estimators. It is also a good entry point into a deeper idea that often appears later in the same paper: unbiasedness alone does not make an estimator good. 2X̄ is unbiased, but it is not the most efficient estimator of θfor this distribution — the maximum order statistic X₍ₙ₎, suitably scaled as ((n+1)/n)·X₍ₙ₎, has much smaller variance and is generally preferred in practice. Recognising that unbiasedness and efficiency are different properties, each with its own criterion, is exactly the kind of distinction that separates a quick right answer from a genuinely solid grasp of estimation theory, which is usually the way this topic is taught in a structured classroom setting at a place like Sunrise Classes rather than through isolated problem-solving.

Frequently Asked Questions

Why is X̄ itself not unbiased for θ when X follows U(0, θ)?

Because the mean of a single Uniform(0, θ) observation is θ/2, not θ. Averaging n such observations does not change this; the sample mean converges to θ/2 as n grows, so X̄ always underestimates θ by a consistent factor of 2. You need to scale it to remove that bias.

Is 2X̄ the best possible estimator of θ?

No. It is unbiased, but it has a relatively large variance compared to estimators built from the sample maximum, such as ((n+1)/n)·X₍ₙ₎. In fact the sample maximum alone, suitably adjusted, is far more efficient because it uses the information that no observation can exceed θ.

Why doesn't taking the square root of an unbiased statistic give another unbiased statistic?

Expectation only passes cleanly through linear operations (addition and multiplication by constants). For a nonlinear, concave function like the square root, Jensen's inequality tells us E(√X̄) is strictly less than √(E(X̄)), so biasedness creeps in even if the original statistic was unbiased.

How do I quickly recognise that a question wants an unbiased estimator rather than the MLE or MVUE?

Look at the phrasing. Unbiased estimator only requires E(T) = θ; it does not ask for the smallest variance or the likelihood-maximising value. If a question instead says most efficient, minimum variance, or maximum likelihood estimator, the required method and often the answer itself will be different.

Does this technique generalise to other distributions, like Exponential or Normal?

Yes. The exact same three-step approach — find E(single observation) in terms of the parameter, find E(X̄), then scale by the right constant — works for any distribution where the parameter enters linearly into the mean. For distributions where the parameter affects variance instead of the mean, you would instead work with E(S²) and scale that.

Why did options (a) and (b) use X̄ and X̄/2 as distractors?

These are classic half-as-much, twice-as-much distractors designed to catch candidates who either forget to scale at all or scale in the wrong direction. Carefully writing out E(X̄) = θ/2 before looking at the options is the surest way to avoid falling for either one.

If anything in this derivation felt unclear, or if you solved it a different way, drop a comment below — and if you found this useful, consider sharing it with a fellow ISS aspirant who is working through the same paper.